Sankalp NEET Full Test-3 Question-43 Solution

Question: 43. The intensity of transmitted light when a polaroid sheet, placed between two crossed polarization at 22.5^{\circ} from the polarization axis of one of the polaroid, is ( I_{0} is the intensity or polarised light after passing through the first polaroid):

(1) \frac{I_{0}}{2}

(2) \frac{I_{0}}{4}

(3) \frac{I_{0}}{8}

(4) \frac{I_{0}}{16}

Answer: Option (3)

Explanation:

The first polaroid produces plane polarized light of intensity I_{0}.

The second and third polaroids are crossed, meaning their transmission axes are perpendicular.

The middle polaroid is placed at an angle 22.5^{\circ} with respect to the polarization axis of the first polaroid.

According to Malus’ law, when light of intensity I passes through a polaroid making an angle \theta with the polarization direction, the transmitted intensity is:

I = I_{0} \cos^{2} \theta

First, light of intensity I_{0} passes through the middle polaroid at angle 22.5^{\circ}:

I_{1} = I_{0} \cos^{2} 22.5^{\circ}

The angle between the middle polaroid and the third polaroid is 90^{\circ} - 22.5^{\circ} = 67.5^{\circ}, so using Malus’ law again:

I = I_{1} \cos^{2} 67.5^{\circ}

Substituting I_{1}:

I = I_{0} \cos^{2} 22.5^{\circ} \cos^{2} 67.5^{\circ}

Using trigonometric values:

\cos^{2} 22.5^{\circ} = \frac{1+\cos 45^{\circ}}{2} = \frac{1+\frac{1}{\sqrt{2}}}{2} \cos^{2} 67.5^{\circ} = \frac{1-\cos 45^{\circ}}{2} = \frac{1-\frac{1}{\sqrt{2}}}{2}

Multiplying the two expressions:

\cos^{2} 22.5^{\circ} \cos^{2} 67.5^{\circ} = \frac{1}{8}

Thus, the transmitted intensity is:

I = \frac{I_{0}}{8}

Hence, the correct answer is Option (3).